LeetCode #3537 — MEDIUM

Fill a Special Grid

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a non-negative integer n representing a 2ⁿ x 2ⁿ grid. You must fill the grid with integers from 0 to 2²ⁿ - 1 to make it special. A grid is special if it satisfies all the following conditions:

All numbers in the top-right quadrant are smaller than those in the bottom-right quadrant.
All numbers in the bottom-right quadrant are smaller than those in the bottom-left quadrant.
All numbers in the bottom-left quadrant are smaller than those in the top-left quadrant.
Each of its quadrants is also a special grid.

Return the special 2ⁿ x 2ⁿ grid.

Note: Any 1x1 grid is special.

Example 1:

Input: n = 0

Output: [[0]]

Explanation:

The only number that can be placed is 0, and there is only one possible position in the grid.

Example 2:

Input: n = 1

Output: [[3,0],[2,1]]

Explanation:

The numbers in each quadrant are:

Top-right: 0
Bottom-right: 1
Bottom-left: 2
Top-left: 3

Since 0 < 1 < 2 < 3, this satisfies the given constraints.

Example 3:

Input: n = 2

Output: [[15,12,3,0],[14,13,2,1],[11,8,7,4],[10,9,6,5]]

Explanation:

The numbers in each quadrant are:

Top-right: 3, 0, 2, 1
Bottom-right: 7, 4, 6, 5
Bottom-left: 11, 8, 10, 9
Top-left: 15, 12, 14, 13
max(3, 0, 2, 1) < min(7, 4, 6, 5)
max(7, 4, 6, 5) < min(11, 8, 10, 9)
max(11, 8, 10, 9) < min(15, 12, 14, 13)

This satisfies the first three requirements. Additionally, each quadrant is also a special grid. Thus, this is a special grid.

Constraints:

0 <= n <= 10

Step 01

Brute Force Baseline

Problem summary: You are given a non-negative integer n representing a 2n x 2n grid. You must fill the grid with integers from 0 to 22n - 1 to make it special. A grid is special if it satisfies all the following conditions: All numbers in the top-right quadrant are smaller than those in the bottom-right quadrant. All numbers in the bottom-right quadrant are smaller than those in the bottom-left quadrant. All numbers in the bottom-left quadrant are smaller than those in the top-left quadrant. Each of its quadrants is also a special grid. Return the special 2n x 2n grid. Note: Any 1x1 grid is special.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

Example 2

Example 3

Step 02

Core Insight

What unlocks the optimal approach

Solve the problem recursively.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3537: Fill a Special Grid
class Solution {
  public int[][] specialGrid(int n) {
    final int sz = 1 << n;
    int[][] grid = new int[sz][sz];
    fill(grid, 0, sz, 0, sz);
    return grid;
  }

  private int count = 0;

  private void fill(int[][] grid, int x1, int x2, int y1, int y2) {
    if (x2 - x1 == 1) {
      grid[x1][y1] = count++;
      return;
    }
    int midRow = (x1 + x2) / 2;
    int midCol = (y1 + y2) / 2;
    fill(grid, x1, midRow, midCol, y2);
    fill(grid, midRow, x2, midCol, y2);
    fill(grid, midRow, x2, y1, midCol);
    fill(grid, x1, midRow, y1, midCol);
  }
}

// Accepted solution for LeetCode #3537: Fill a Special Grid
package main

// https://space.bilibili.com/206214
func specialGrid(n int) [][]int {
	val := 0
	var dfs func([][]int, int)
	dfs = func(a [][]int, l int) {
		if len(a) == 1 {
			a[0][l] = val
			val++
			return
		}
		m := len(a) / 2
		dfs(a[:m], l+m)
		dfs(a[m:], l+m)
		dfs(a[m:], l)
		dfs(a[:m], l)
	}

	a := make([][]int, 1<<n)
	for i := range a {
		a[i] = make([]int, 1<<n)
	}
	dfs(a, 0)
	return a
}

func specialGrid2(n int) [][]int {
	a := make([][]int, 1<<n)
	for i := range a {
		a[i] = make([]int, 1<<n)
	}
	val := 0
	var dfs func(int, int, int, int)
	dfs = func(u, d, l, r int) {
		if d-u == 1 {
			a[u][l] = val
			val++
			return
		}
		m := (d - u) / 2
		dfs(u, u+m, l+m, r)
		dfs(u+m, d, l+m, r)
		dfs(u+m, d, l, l+m)
		dfs(u, u+m, l, l+m)
	}
	dfs(0, 1<<n, 0, 1<<n)
	return a
}

# Accepted solution for LeetCode #3537: Fill a Special Grid
class Solution:
  def specialGrid(self, n: int) -> list[list[int]]:
    sz = 1 << n
    grid = [[0] * sz for _ in range(sz)]
    count = 0

    def fill(x1: int, x2: int, y1: int, y2: int) -> None:
      nonlocal count
      if x2 - x1 == 1:
        grid[x1][y1] = count
        count += 1
        return
      midRow = (x1 + x2) // 2
      midCol = (y1 + y2) // 2
      fill(x1, midRow, midCol, y2)
      fill(midRow, x2, midCol, y2)
      fill(midRow, x2, y1, midCol)
      fill(x1, midRow, y1, midCol)

    fill(0, sz, 0, sz)
    return grid

// Accepted solution for LeetCode #3537: Fill a Special Grid
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3537: Fill a Special Grid
// class Solution {
//   public int[][] specialGrid(int n) {
//     final int sz = 1 << n;
//     int[][] grid = new int[sz][sz];
//     fill(grid, 0, sz, 0, sz);
//     return grid;
//   }
// 
//   private int count = 0;
// 
//   private void fill(int[][] grid, int x1, int x2, int y1, int y2) {
//     if (x2 - x1 == 1) {
//       grid[x1][y1] = count++;
//       return;
//     }
//     int midRow = (x1 + x2) / 2;
//     int midCol = (y1 + y2) / 2;
//     fill(grid, x1, midRow, midCol, y2);
//     fill(grid, midRow, x2, midCol, y2);
//     fill(grid, midRow, x2, y1, midCol);
//     fill(grid, x1, midRow, y1, midCol);
//   }
// }

// Accepted solution for LeetCode #3537: Fill a Special Grid
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3537: Fill a Special Grid
// class Solution {
//   public int[][] specialGrid(int n) {
//     final int sz = 1 << n;
//     int[][] grid = new int[sz][sz];
//     fill(grid, 0, sz, 0, sz);
//     return grid;
//   }
// 
//   private int count = 0;
// 
//   private void fill(int[][] grid, int x1, int x2, int y1, int y2) {
//     if (x2 - x1 == 1) {
//       grid[x1][y1] = count++;
//       return;
//     }
//     int midRow = (x1 + x2) / 2;
//     int midCol = (y1 + y2) / 2;
//     fill(grid, x1, midRow, midCol, y2);
//     fill(grid, midRow, x2, midCol, y2);
//     fill(grid, midRow, x2, y1, midCol);
//     fill(grid, x1, midRow, y1, midCol);
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.